Given an index k, return the kth row of the Pascal‘s triangle.
For example, given k = 3,
Return [1,3,3,1].
Note:
Could you optimize your algorithm to use only O(k) extra space?
这道题是要在O(k)个空间内，找到第k行数据。如果直接顺序计算a[i]=a[i]+[i-1],由于a[i]被替换，所以，在计算a[i+1]时候就出错了。所以使用倒序计算
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<int> a;
a.resize(rowIndex+1,0);
a[0]=1;
for(int k =0;k <rowIndex +1;k++)
{
for(int i =k;i>0;i--)
{
a[i]=a[i-1]+a[i];
}
}
return a;
}
};
(leetcode)Pascal's Triangle II
原文：http://www.cnblogs.com/chdxiaoming/p/4478438.html